Proof.1o¯superscript1normal-¯o1^{{\underline{o}}}. Suppose that f:A→Bnormal-:fnormal-→ABf\!:\,A\to B is surjective. Let XXX be an arbitrary subset of AAA and yyy any element of the setB∖f⁢(X)BfXB\!\smallsetminus\!f(X). By the surjectivity, there is an xxx in AAA such that f⁢(x)=yfxyf(x)=y, and since y∉f⁢(X)yfXy\notin f(X), the element xxx is not in XXX, i.e. x∈A∖XxAXx\in A\!\smallsetminus\!X and thus y=f⁢(x)∈f⁢(A∖X)yfxfAXy=f(x)\in f(A\!\smallsetminus\!X). One can conclude that B∖f⁢(X)⊆f⁢(A∖X)BfXfAXB\!\smallsetminus\!f(X)\,\subseteq\,f(A\!\smallsetminus\!X) for all X⊆AXAX\subseteq A.

2o¯superscript2normal-¯o2^{{\underline{o}}}. Conversely, suppose the condition (1). Let again XXX be an arbitrary subset of AAA and yyy any element of BBB. We have two possibilities:
a) y∉f⁢(X)yfXy\notin f(X); then y∈B∖f⁢(X)yBfXy\in B\!\smallsetminus\!f(X), and by (1), y∈f⁢(A∖X)yfAXy\in f(A\!\smallsetminus\!X). This means that there exists an element xxx of A∖X⊆AAXAA\!\smallsetminus\!X\subseteq A such that f⁢(x)=yfxyf(x)=y.
b) y∈f⁢(X)yfXy\in f(X); then there exists an x∈X⊆AxXAx\in X\subseteq A such that f⁢(x)=yfxyf(x)=y.
The both cases show the surjectivity of fff.